Flux across a curve | MIT 18.02SC Multivariable Calculus, Fall 2010

[@TerresEndormies]

The curve is closed, so I wouldn't use an inward-pointing N. I would use N = (x,y) which HAPPENS to be the given vector field. That's the point. Using geometric methods helps clarify relationships that can be confusing if you churn out an algebraic formula. But of course feel free to use the in-pointer. Try un-parameterizing the equation. What's the gradient of 0 = F = x^2 + y^2 - 1 ? Does that help?

[@jamaljaffer8412]

Brilliant explanation and many thanks

[@samas69420]

is true that flux (suppose that flux is only in term of the radius, flux=f(r)) computed through any closed curve in the plane is equal to the flux computed through the unit circle?

[@Archmage5023]

Wonderful explanation, thanks

[@mritunjaybhatt9277]

Thank you mam.....I learnt a lot

[@matmicmj]

PEACE AND LOVE .,.,.,.,

[@kinslay3r]

I have no idea what the flux is going on in this video.

[@bobkameron]

haha

[@imegatrone]

I Really Like The Video Flux across a curve From Your

[@TerresEndormies]

giggle.... I want more naughty language in math videos. Orienting objects with the signs of sine and cosine parameterizations, under change of coordinate space, is like my personal enemy. I find it relentlessly confusing. So, I hope my comment wasn't even more puzzling. I'm trying to work this stuff out alone; commenting helps me test myself. Let me know if I'm wrong!

[@npa50051]

I used to like math... then i took these classes

[@AlphaBeta-xt3wn]

WOULD

[@nikolaromanos456]

cut doods + flux